Відповіді АЛГЕБРА: 15.39 (1)

15.39

1) cos²x + cos²y – cos(x – y) • cos(x + y) = cos²x + cos²y – 12(cos(x –

– y – x – y) + cos(x – y + x + y)) = cos²x + cos²y – 12cos(–2y)– 12cos(2x) = cos²x – 12 cos²x + cos²y – 12cos2y = cos²x – 12(cos²x – sin²x) + cos²y – 12(cos²y – sin²y) = 12cos²x + 12sin²x + 12cos²y + 12sin²y = 12 + 12 = 1.

2) sin²(α – 3π8) – cos²(α + 3π8) = cos²(π\42 – α + 3π8) – cos²(α + 3π8) =

= cos²(π8 – α) – cos²(α + 3π8) = (cos(π8 – α) – cos(α + 3π8))(cos(π8 – α) + cos(α + 3π8)) = –2sinπ8-α+α+3π82 sin π8-α-α-3π82 • 2cosπ8-α+α+3π82 •

• cosπ8-α-α-3π82 = –4 • sinπ4 • sin(–π8 – α) • cosπ4 • cos(–π8 – α) =

= –4 • 22 • (–sin(π4 + α)) • 22 • cos(π4 + α) = 2sin(π4 + α)cos(π4 + α) = sin(π2 + 2α) = cos2α.

3) (1sin3α + 1sinα) • (sin4αsinα – cos4αcosα) = sinα+sin3αsin3αsinα •

• sin4αcosα-cos4αsinαsinαcosα = 2sin2αcosα • sin3αsin3αsinα • sinαcosα = 2sin2αsin2α =

= 4sinαcosαsin2α = 4ctgα

4) cosα-cos3α(sinα+sin3α)1-cos4α = -2sin2αsinα • 2sin2αcosα2sin22α =

= -2sin22α sinαcosαsin22α = –sin2α